Two point white dots are 1 mm apart on a black paper. They are viewed by eye of pupil diameter 3 mm.

Q: Why is there a limit to how far we can see two dots? Shouldn't we always see them if they are bright enough?

It's not about brightness, but about diffraction. Because light is a wave, it bends when passing through your pupil, creating a blurred spot (Airy disk) on your retina instead of a perfect point. As you move further away, the angle between the dots gets smaller, and their blurred spots overlap. When they overlap too much, your brain can't tell them apart.

Q: Where does the $1.22$ factor come from in the formula $\theta_R = 1.22 \lambda / d$?

The factor $1.22$ comes from the complex mathematics of diffraction through a circular aperture. Specifically, it is the first root of a Bessel function of the first kind. It represents the exact angle at which the first dark ring of the diffraction pattern occurs.

Q: How does the wavelength of light affect our ability to resolve objects?

The limit of resolution $\theta_R$ is directly proportional to the wavelength $\lambda$. This means shorter wavelengths (like blue or violet light) produce a smaller $\theta_R$, allowing for better resolution. This is why electron microscopes, which use electrons with incredibly short wavelengths, can resolve much smaller details than optical microscopes.

Q: Would having a larger pupil diameter help us see the dots from further away?

Yes, absolutely! The limit of resolution $\theta_R$ is inversely proportional to the pupil diameter $d$. A larger pupil means a smaller $\theta_R$, which translates to better resolving power. This is the primary reason why astronomical telescopes are built with massive objective lenses or mirrors—to gather more light and to achieve incredibly high resolving power.

Q: The calculation gave $4.91\text{ m}$, but the answer is $5\text{ m}$. Why did we approximate?

In many competitive exams like JEE, options are sometimes rounded to the nearest integer or significant figure to test your conceptual understanding rather than exact arithmetic precision. Since $4.91\text{ m}$ is very close to $5\text{ m}$ and far from the other options ($1\text{ m}, 3\text{ m}, 6\text{ m}$), $5\text{ m}$ is the unambiguously correct choice.

Question

Two point white dots are $1	ext{ mm}$ apart on a black paper. They are viewed by eye of pupil diameter $3	ext{ mm}$. Approximately, what is the maximum distance at which these dots can be resolved by the eye?
[Take wavelength of light $= 500	ext{ nm}$]

Options:

(A) $5	ext{ m}$

(B) $1	ext{ m}$

(C) $6	ext{ m}$

(D) $3	ext{ m}$

Sigma Percentile · Accepted Answer

Have you ever looked at a car approaching from far away at night? Initially, its two headlights look like a single, bright blob of light. But as the car gets closer, that single blob magically splits into two distinct headlights. This everyday phenomenon is a perfect demonstration of the **resolving power** of your eyes.

In this problem, we are asked to find the maximum distance at which our eyes can distinguish two tiny white dots separated by just $1	ext{ mm}$. Let's dive into the fascinating physics of why our vision has limits and how we can calculate them.

### The Physics: Diffraction and Rayleigh's Criterion

When light from an object enters our eye, it passes through the pupil, which acts as a circular aperture. Because light behaves as a wave, it doesn't just travel in straight lines; it bends or **diffracts** around the edges of the pupil. Instead of forming a perfect point on our retina, a point source of light forms a central bright spot surrounded by faint, concentric rings. This pattern is known as an **Airy disk**.

When we look at two dots, each dot forms its own Airy disk on our retina. If the dots are far away, the angle they subtend at our eye is very small, and their Airy disks overlap significantly. If they overlap too much, our brain interprets them as a single object.

Lord Rayleigh proposed a criterion to determine when two objects are "just resolved." **Rayleigh's Criterion** states that two point sources are just resolved when the center of the Airy disk of one source falls exactly on the first dark ring (minimum) of the Airy disk of the second source.

Mathematically, for a circular aperture of diameter $d$, the minimum angular separation $	heta_R$ (the limit of resolution) is given by:

$$ 	heta_R = \frac{1.22 \lambda}{d} $$

where $\lambda$ is the wavelength of the light.

### The Setup: Dots and the Eye

Let's translate this to our specific problem. We have two dots separated by a distance $y = 1	ext{ mm}$. We are viewing them from a distance $D$. The angle $	heta$ that these two dots subtend at our eye can be approximated using simple geometry (since the angle is very small):

$$ 	heta \approx \frac{y}{D} $$

For our eye to resolve these two dots, the angle they subtend must be greater than or equal to the eye's limit of resolution:

$$ 	heta \ge 	heta_R $$

### The Master Equation

Substituting our expressions for $	heta$ and $	heta_R$, we get the master equation for this problem:

$$ \frac{y}{D} \ge \frac{1.22 \lambda}{d} $$

We want to find the maximum distance, $D_{	ext{max}}$. Rearranging the inequality to solve for $D$, we get:

$$ D \le \frac{y \cdot d}{1.22 \lambda} $$

Therefore, the maximum distance is exactly when the equality holds:

$$ D_{	ext{max}} = \frac{y \cdot d}{1.22 \lambda} $$

### The Final Calculation

Now, it's time to plug in the numbers. It is absolutely critical to convert all given values into standard SI units (meters) to avoid silly mistakes.

*   Separation between dots, $y = 1	ext{ mm} = 10^{-3}	ext{ m}$
*   Pupil diameter, $d = 3	ext{ mm} = 3 	imes 10^{-3}	ext{ m}$
*   Wavelength of light, $\lambda = 500	ext{ nm} = 500 	imes 10^{-9}	ext{ m} = 5 	imes 10^{-7}	ext{ m}$

Substituting these into our equation for $D_{	ext{max}}$:

$$ D_{	ext{max}} = \frac{(10^{-3}	ext{ m}) 	imes (3 	imes 10^{-3}	ext{ m})}{1.22 	imes (5 	imes 10^{-7}	ext{ m})} $$

$$ D_{	ext{max}} = \frac{3 	imes 10^{-6}}{6.1 	imes 10^{-7}} $$

$$ D_{	ext{max}} = \frac{30}{6.1} \approx 4.91	ext{ m} $$

Looking at the given options, the closest value is **$5	ext{ m}$**. Thus, if you stand further than 5 meters away, those two dots will blur into one. This beautiful interplay of geometry and wave optics governs not just human vision, but the design of every telescope and microscope in existence!

Animated Solution for Physics - Optics: Two point white dots are $1 mm$ apart on a black paper. They are viewed by eye of pupil diameter $3 mm$ . Approximately, what is the maximum distance at which these dots can be resolved by the eye?\n[Take wavelength of light $= 500 nm$ ]

Select Answer:

Visualized Solution

\text{Visualizing the Setup}

\text{Rayleigh's Criterion}

\text{Formulating the Condition}

\text{Substituting Values}

\text{Calculation}

\text{Conclusion}

The Sigma Insight: Optical Instruments

The Physics

The Setup

The Master Equation

The Final Calculation

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